Question

Consider the following reactions.
\(PbCl_2 + K_2CrO_4 \rightarrow A + 2KCl\) (Hot solution)
\(A + NaOH \rightarrow B + Na_2CrO_4\)
\(PbSO_4 + 4CH_3COONH_4 \rightarrow (NH_4)_2SO_4 + X\)
In the above reactions, A, B and X are respectively:

• A. \(Na_2[Pb(OH)_2]\), \(PbCrO_4\) and \((NH_4)_2[Pb(CH_3COO)_4]\)
• B. \(Na_2[Pb(OH)_2]\), \(PbCrO_4\) and \([Pb(NH_3)_4]SO_4\)
• C. \(PbCrO_4\), \(Na_2[Pb(OH)_4]\) and \([Pb(NH_3)_4]SO_4\)
• D. \(PbCrO_4\), \(Na_2[Pb(OH)_4]\) and \((NH_4)_2[Pb(CH_3COO)_4]\)

Hint:

PbSO4 is soluble in ammonium acetate. This is a distinguishing test for Pb in qualitative analysis.

Answer 1

The Correct Option is D

To solve the problem, let's analyze the reactions step-by-step:

1. First Reaction: PbCl_2 + K_2CrO_4 \rightarrow A + 2KCl (Hot solution)
   • Lead chloride reacts with potassium chromate to form lead chromate and potassium chloride.
   • Here, A is PbCrO_4, which is a yellow precipitate commonly formed in such reactions.
2. Second Reaction: A + NaOH \rightarrow B + Na_2CrO_4
   • Lead chromate reacts with sodium hydroxide to form sodium plumbate and sodium chromate.
   • Thus, B is Na_2[Pb(OH)_4].
3. Third Reaction: PbSO_4 + 4CH_3COONH_4 \rightarrow (NH_4)_2SO_4 + X
   • Lead sulfate reacts with ammonium acetate to form ammonium sulfate and a complex lead acetate compound.
   • Thus, X is (NH_4)_2[Pb(CH_3COO)_4].

Thus, the substances A, B, and X are PbCrO_4, Na_2[Pb(OH)_4], and (NH_4)_2[Pb(CH_3COO)_4], respectively.

The correct answer is therefore the option PbCrO_4, Na_2[Pb(OH)_4], and (NH_4)_2[Pb(CH_3COO)_4].